It's well known that in low orbits around astronomical bodies with uneven mass distribution, orbits will shift around, and the orbiting object may crash into the surface.

Why is this the case? Newtonian gravity seems to dictate that we can treat all masses as point masses. In this case, a satellite orbiting a planet would simply be two point masses orbiting around their common center of mass. Why would the mass distribution matter at all? The planet, I thought, could even be some really really long cylinder and a satellite would still orbit in perfect ellipses around its center of mass.


It is incorrect to think that

Newtonian gravity seems to dictate that we can treat all masses as point masses.

Indeed, you can't do that, and one of the consequences of that fact is the orbital decay that you refer to.

The theorem you're thinking of holds only for bodies with perfect spherical symmetry, and states that for such a body, the gravitational field outside it is identical to that of a point mass at the body's centre.

However, you run into trouble immediately after you leave that setting. For example, the interactions between two spherical bodies cannot, in general, be reduced to the interactions between point masses. You can indeed substitute one body for a point mass to find its field, but then you need to find the force on the second body caused by this spatially inhomogeneous field.

It is, sometimes but not always, a good approximation to treat this as a total force acting on its centre of mass:

  • If the second body is very small compared to the bodies' separation, then the relative variation of the first body's gravitational field over the second body's spatial extent will probably be small, and it can be neglected. As a better approximation, you can include the first-order variation of the first body's gravitational field, which is called a tidal force.

  • If the second body is perfectly rigid, it will respond to any inhomogeneous force field by means of displacements of the COM caused by the total force, which is again the point-particle model. As a better approximation, you can include linear deformations caused by first-order variations in the force field. If you couple this with the tidal forces, you'll have a nice basic description of tidal drag.

Notice, for example, that the Moon obeys both conditions pretty well.

For a body with an arbitrary mass distribution, you can probably now guess that it gets a lot hairier. If the distribution is really wild, then there is nothing for it but to calculate the field numerically, and orbits will be hard to come by. If the body is mostly spherical, then there is some hope.

The Earth and the Moon are sort of on the latter case, but their distributions are far from uniform. The Earth's geoid, which describes its shape as far as gravity goes, is not spherical:

enter image description here

Image source. Note the exaggerated scale: 50 mGal corresponds to a change in $g$ of 0.005%.

For the Moon, mass concentrations underneath mare make the gravity even more uneven.

The tool to deal with these departures from sphericality is borrowed from electrostatics: the multipole expansion. That is, if the body is mostly spherical, then you can treat its gravitational field as a superposition of a point mass plus a point dipole, quadrupole, octupole, and so on.

$$\Phi(\mathbf r)=\sum_{l,m}Q_{lm} \frac{Y_l^m(\theta, \phi)}{r^{l+1}}.$$

The good news is that the contribution of each succeeding term gets weaker and weaker, and most importantly they get weaker the further away the test mass is. As a bonus, you can usually make the dipolar terms to zero by putting the origin at the centre of mass.

For the Earth, the first non-point-mass term that you need to account for is the zonal quadrupole term, with $l=2$ and $m=0$, which describes the Earth's bulge at the equator and the corresponding flatness at the poles. Including that as in this article, you get a gravitational potential that looks like $$ \Phi=-\frac{GM}{r}\left[1-J_2 \left(\frac{a_1}{r}\right)^2P_2(\cos\theta)\right]. $$ Here $a_1$ is the Earth's equatorial radius, and $J_2\approx0.001$ is called the quadrupole moment coefficient. For low Earth orbit, this effect is on the order of 0.1%, which is small but it's a perturbation that needs to be dealt with. For orbits which are further than 3 or 4 $a_1$ from Earth, the effect becomes correspondingly smaller, in a quadratic fashion. This is on the order of 0.01% of the Keplerian force, but again it's a perturbation that needs to be accounted for and which can add up over many orbits to give sizeable effects.

  • 2
    $\begingroup$ $J_2$ is typically called the Earth's second dynamic form factor. The quadrupole moment is slightly different. There's a lot more to the Earth's gravity field than just the dynamic form factor $J_2$. The images you posted are of gravity anomalies, the deviations of the Earth's gravity field from that generated by an oblate spheroid. The GRACE GGM03C Gravity Model is a 360x360 spherical harmonic model. An even larger model, the EGM2008, is a 2159x2159 model with extensions to 2190x2159. $\endgroup$ – David Hammen Sep 29 '14 at 6:13

Newtonian gravity seems to dictate that we can treat all masses as point masses.

This is the source of your confusion. This is not true. One can treat an object with a spherically symmetric mass distribution (i.e., density is a function of distance from the center of the object) as if were a point mass for points outside of the object in question. There is no error in Newtonian mechanics from such a treatment, so long as the object does indeed have a spherically symmetric mass distribution.

For objects that do not have a spherically symmetric mass distribution (e.g., the planets and their moons), the point mass approximation is just that, an approximation. The approximation is very good at very large distances. It's not so good at close distances.

The Earth and our Moon provide a couple of close-to-home examples. With regard to the Earth's Moon, I suggest you read about the fate of a pair of satellites sent into orbit about the Moon by the Apollo program, PFS-2 and PFS-1. You can read about their bizarre fate here: http://science.nasa.gov/science-news/science-at-nasa/2006/06nov_loworbit/.

With regard to the Earth, the Earth has a marked equatorial bulge thanks to its rotation. There are a good number of satellites in low Earth orbit whose orbits are explicitly designed to take advantage of the Earth's equatorial bulge. These are the Sun-synchronous satellites. The equatorial bulge causes orbits to precess (Keplerian orbits do not precess). Set up the orbit just right and the precession will be in synchronization with the Earth's orbit about the Sun. These Sun synchronous orbits are very important for satellites that observe the Earth.

That those Sun synchronous orbits exist are something that satellite designers take advantage of. The non-spherical nature of the Earth presents challenges to designers of satellites that aren't supposed to precess. For example, it would be very nice if geostationary satellites were in fact geostationary. That is not the case. A second order effect of the Earth's non-spherical mass distribution makes geostationary satellites tend to migrate toward the longitude of either India or California. Geostationary satellite designers need to provide fuel to counteracts that tendency to move to those preferred longitudes.

  • $\begingroup$ "One can treat an object with a spherically symmetric mass distribution...". I am afraid that such an approximation is not even valid for a spherically symmetric body that rotates. Landau/Lifshitz make that a central point in the second sentence of their volume about mechanics, or so. $\endgroup$ – CuriousOne Sep 29 '14 at 0:44
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    $\begingroup$ @CuriousOne - That's correct in general relativity. It's the Lense-Thirring effect. In Newtonian mechanics, which is the subject of this question, what you wrote is incorrect. There is no Newtonian analog of the Lense-Thirring effect. $\endgroup$ – David Hammen Sep 29 '14 at 5:55
  • $\begingroup$ I cite Landau/Lifshitz Volume I Classical Mechanics: "Paragraph 1. Generalized Coordinates. One of the fundamental concepts of mechanics is that of a particle. By this we mean a body whose dimensions may be neglected in describing its motion. The possibility of doing so depends, of course, on the conditions of the problem concerned. For example, the planets may be regarded as particles in considering their motion about the sun, but not considering their rotation about their axes.". Would you agree about the physics of the rotating solid body being non-trivial? $\endgroup$ – CuriousOne Sep 29 '14 at 18:47
  • $\begingroup$ @CuriousOne - Of course the dynamics of a rotating solid body is non-trivial. I've spent multiple years of work on this very subject. However, the Newtonian gravitational effects of a solid body with a symmetrically spherical mass distribution versus those of a point mass with the same mass are indistinguishable, regardless of the rotation of the non-point mass object. Do the math. $\endgroup$ – David Hammen Sep 29 '14 at 19:04
  • $\begingroup$ The point is that the effects on motion under gravity (or any other central force) IS NOT the correct criterion for the switch from a center of mass dynamic to a solid (or non-solid) body dynamic. The correct criterion is whether internal degrees of freedom are excited. That, by the way, also holds for orbital decay, because if there is no conversion to heat inside the asymmetric body, then there can be no orbital decay because of energy conservation, either. There would be a slight change in orbital shape, though. $\endgroup$ – CuriousOne Sep 29 '14 at 19:21

Point masses are an approximation that describe the motion of the center of mass of objects which do not have their rotational and internal vibrational degrees of freedom excited. If that assumption is incorrect, then so is the treatment of Newtonian problems by using coordinates of centers of mass alone. Orbital decay can only occur if both energy is lost AND angular momentum is being transferred. Energy loss requires that some non-elastic internal degree of freedom converts orbital energy into heat (or some other form of internal energy) and angular momentum transfer can only occur between orbital angular momentum and rotational angular momentum.

The Moon, for instance, is increasing its distance to Earth, because rotational angular momentum of Earth is converted into orbital angular momentum of the Moon. In return the Moon loses energy, which drives the tides in Earth's oceans and crust. As a consequence the planet's rotation is slowing down. This process would stop when one day has reached the length of one month. Given the weak coupling that process, however, will take longer than the Earth-Moon system will last (because the sun will become a red giant before that happens).

In LEO, of course, the main effect is the atmosphere of Earth, which produces significant drag. That, too, is a transfer of both orbital energy and angular momentum, but it has nothing to do with mass distribution.


Because gravitational force increases with proximity and mass, so areas of higher density and/or elevation exert a greater pull on a satellite, creating what could be called a "speed bump", causing the orbiting body to be slowed/pulled down every time it passes that point, if only a little bit.

Also, when you orbit something you are countering its net gravitational pull, but these "speed bumps" cause the net gravitational pull to rise at least slightly when an orbiting body passes over them, resulting in a decaying orbit.


protected by ACuriousMind May 9 '17 at 14:59

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